Single Number I && II

Change-Id: Iadeff09c004a9b330e715c0b3f9d9ccf72f98848

Author: applewjg

SingleNumber.java

@@ -0,0 +1,23 @@
+/*
+ Author:     King, [email protected]
+ Date:       Jan 3, 2015
+ Problem:    Single Number
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/single-number/
+ Notes:
+ Given an array of integers, every element appears twice except for one. 
+ Find that single one.
+ Your algorithm should have a linear runtime complexity. 
+ Could you implement it without using extra memory?
+
+ Solution: XOR.
+*/
+public class Solution {
+    public int singleNumber(int[] A) {
+        int res = 0;
+        for (int i : A) {
+            res = res^i;
+        }
+        return res;
+    }
+}

SingleNumberII.java

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+/*
+ Author:     King, [email protected]
+ Date:       Jan 3, 2015
+ Problem:    Single Number II
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/single-number-ii/
+ Notes:
+ Given an array of integers, every element appears three times except for one. 
+ Find that single one.
+ Your algorithm should have a linear runtime complexity. Could you implement it 
+ without using extra memory?
+
+ Solution: 1. Count the number of each bit.
+        2. We can improve this based on the previous solution using three bitmask variables.
+        3. An excellent answer by @ranmocy in LeetCode Discuss:
+        https://oj.leetcode.com/discuss/857/constant-space-solution?show=2542#a2542
+*/
+public class Solution {
+    public int singleNumber_1(int[] A) {
+        int res = 0;
+        for (int i = 0; i < 32; ++i) {
+            int one = 0;
+            for (int num : A) {
+                if (((num >> i) & 1) == 1) ++one;
+            }
+            res = res | ((one % 3)<<i);
+        }
+        return res;
+    }
+    public int singleNumber_2(int[] A) {
+        int one = 0, twice = 0;
+        for (int num : A) {
+            twice = twice | (one & num);
+            one = one ^ num;
+            int three = one & twice;
+            one = one ^ three;
+            twice = twice ^ three;
+        }
+        return one;
+    }
+    public int singleNumber(int[] A) {
+        int k = 1, n = 3;
+        int[] x = new int[n];
+        x[0] = ~0;
+        for (int num : A) {
+            int t = x[n-1];
+            for (int i = n - 1; i >= 1; --i) {
+                x[i] = (x[i-1] & num) | (x[i] & ~num);
+            }
+            x[0] = (t & num) | (x[0] & ~num);
+        }
+        return x[k];
+    }
+}